Find the acceleration and tension in pulley setup

Question

Could anyone help me out with the following please.

I am trying to find the the acceleration of A and the tension in the strings in the system in the following picture:

I'll call the 3kg mass B and the movable pulley C and the 5kg mass D.

I'll call the tension in the upper string $T_1$ and the tension in the lower string $T_2$.

Now, let's say that D moves down with an acceleration $a_1$ then C must move up with the same.

If C were fixed then B would move down with acceleration say $a_2$ and A would move up with the same.

So acceleration of B will be $a_2 - a_1$ and that of A will be $a_2 + a_1$ and so applying F=ma to all gives me:

for D: $5g-T_1=5a_1$

for C: $T_1 - 2T_2=0$ (as mass of C is zero)

for A: $T_2 - 2g=2(a_1 + a_2)$

for B: $3g - T_2=5(a_2 - a_1)$

Well, if I solve these for $T_2$ I get $T_2 = \frac{36g}{15}$

But book answer is $\frac{120g}{49}$

So I have set up my equations wrongly somehow.

Thanks for any help,

Mitch

Answer 1

I checked your arithmetic and your equations. Your arithmetic is OK, but your equation for the acceleration of $B$ should read: $$3g-T_2=3(a_2-a_1)$$ Because the mass of block $B$ is $3$ kg, not $5$ kg! Making this change and solving the equations the same way, I do indeed get $T_2=\frac{120kg\cdot g}{49}$, so that is my diagnosis of your error.